SSC CGL 23 September 2025 · Shift 3 · Quantitative Aptitude — questions with answers · ShikshaSphere
Chapter 143 of 192
23 September 2025 · Shift 3
Quantitative Aptitude
25 questions ~25 min readFree
25 questions
A
51
A sum of money triples itself at compound interest in 10 years. In how many years will it become nine times?
A.20
B.25
C.30
D.15
Solution
CI — multiples
Under compound interest the multiple grows in powers.
9 = 3², so the tripling must happen twice over.
Time = 2 × 10.
= 20 years — option (a).
52
Which of these numbers is neither Rational nor an Integer?
A.
B.−5
C.0
D.4.5
Solution
Irrational number
A rational number can be written as ; −5, 0 and 4.5 all can.
= 1.732050… — its decimal neither ends nor repeats.
So it cannot be written as such a fraction and is irrational.
Hence option (a).
53
A train starts from Mumbai at 6 a.m. and arrives at Kolhapur at 2:30 p.m. on the same day. If the speed of the train is 120 km per hour, find the distance travelled by the train.
A.900 km
B.980 km
C.1020 km
D.1000 km
Solution
Distance from time
From 6:00 a.m. to 2:30 p.m. is 8 hours 30 minutes = 8.5 hours.
Distance = speed × time.
= 120 × 8.5.
= 1020 km — option (c).
54
A spherical tank is filled with water. The radius of the sphere is 12 metres. The tank is used to fill cylindrical containers, each with a radius of 2 metres and a height of 24 metres. How many containers can be filled?
A perfect cube with a side length of 4 cm has the largest possible sphere fitted inside it. What is the volume of the empty space within the cube?
A.64 − cm³
B.12 − cm³
C.64 − cm³
D. cm³
Solution
Cube minus sphere
Cube volume = 4³ = 64 cm³.
The largest sphere has diameter equal to the edge, so r = 2 cm.
Sphere volume = π(8) = cm³.
Empty space = 64 − cm³ — option (a).
56
'A' and 'B' entered a joint venture. 'A' brought ₹3,00,000 and 'B' ₹4,00,000. 'A' also contributed skilled labour equivalent to ₹1,00,000. At the end of the year they earned ₹2,40,000 in profit. If they agreed to divide the profit in proportion to their effective contributions, what is A's share?
A.₹1,10,000
B.₹1,20,000
C.₹1,24,500
D.₹1,50,000
Solution
Effective contributions
A’s effective contribution = 3,00,000 + 1,00,000 = ₹4,00,000.
B’s contribution = ₹4,00,000.
So the ratio is 4,00,000 : 4,00,000 = 1 : 1.
A’s share = = ₹1,20,000 — option (b).
57
A banner is shaped like a trapezium with parallel sides 10 m and 6 m, and height 8 m. If printing costs ₹500 per m², what is the total cost?
A.₹32,000
B.₹30,000
C.₹28,000
D.₹25,000
Solution
Trapezium area
Area of a trapezium = × (sum of parallel sides) × height.
= × (10 + 6) × 8 = 64 m².
Cost = 64 × 500.
= ₹32,000 — option (a).
58
A farmer's land consists of two adjoining trapezoids: T₁ with parallel sides 35 m and 25 m and height 10 m; T₂ with parallel sides 25 m and 15 m and height 10 m. Land costs ₹30 per m². What is the total land cost?
A.₹15,000
B.₹30,000
C.₹24,000
D.₹25,000
Solution
Two trapezoids
T₁ = × (35 + 25) × 10 = 300 m².
T₂ = × (25 + 15) × 10 = 200 m².
Total area = 300 + 200 = 500 m².
Cost = 500 × 30 = ₹15,000 — option (a).
59
The ratio of areas of two regular polygons with the same number of sides is 16 : 25. What is the ratio of their side lengths?
A.5 : 4
B.4 : 5
C.2 : 3
D.4 :
Solution
Areas and sides
For similar figures, area varies as the SQUARE of the corresponding length.
So the side ratio = .
= : .
= 4 : 5 — option (b).
60
A student scored 40% in one subject and 30% in another. Both subjects have equal maximum marks. What is the overall percentage?
A.30%
B.35%
C.32%
D.38%
Solution
Overall percentage
Since the maximum marks are equal, we can take each as 100.
Marks obtained = 40 + 30 = 70 out of 200.
× 100.
= 35% — option (b).
61
A seller offers a combo deal: "Buy 4 shirts for the price of 3". If each shirt costs him ₹1,500, what is the effective loss percentage?
A.20%
B.25%
C.30%
D.33%
Solution
Buy 4 pay for 3
Cost of 4 shirts = 4 × 1,500 = ₹6,000.
He receives the price of only 3 = 3 × 1,500 = ₹4,500.
Loss = 6,000 − 4,500 = ₹1,500.
Loss% = × 100 = 25% — option (b).
62
The price of a product is decreased by y% and then increased by y%. The final price becomes 10% less than the original. What is the value of y (approximately)?
A.31.6
B.20.6
C.25.6
D.29.6
Solution
Fall then rise
A fall of y% followed by a rise of y% leaves (1 − )(1 + ) = 1 − .
This must equal 0.9, so = 0.1.
y² = 1000.
y = ≈ 31.6 — option (a).
63
The average of 6 different numbers is 120. If the smallest number is 18, what is the average of the remaining five numbers?
A.127.5
B.140.4
C.125.5
D.124.5
Solution
Average of the rest
Total of the 6 numbers = 6 × 120 = 720.
Removing 18 leaves 720 − 18 = 702.
Average of the remaining five = .
= 140.4 — option (b).
64
A prism with a parallelogram base has base area 160 cm² and height 24 cm. If its lateral surface area is 768 cm², what is the perimeter of the base?
A.26 cm
B.22 cm
C.36 cm
D.32 cm
Solution
Lateral surface area
For any prism, lateral surface area = perimeter of base × height.
768 = perimeter × 24.
Perimeter = .
= 32 cm — option (d).
65
The average score of 50 students in a test is 136. The top 5 scorers had an average of 160 and the lowest 5 had an average of 100. What is the average of the remaining 40 students?
A.158
B.137.5
C.117
D.216
Solution
Average of the middle group
Total of all 50 = 50 × 136 = 6800.
Top 5 total = 5 × 160 = 800; lowest 5 total = 5 × 100 = 500.
Remaining 40 total = 6800 − 800 − 500 = 5500.
Average = = 137.5 — option (b).
66
A wholesaler offers a bulk order deal where a 20% discount is applied to orders above ₹10,000, and a 5% additional seasonal discount is available. What is the effective price paid for an order worth ₹50,000?
A.₹34,100
B.₹25,500
C.₹30,000
D.₹38,000
Solution
Successive discounts
After the 20% discount: 50,000 × = ₹40,000.
The 5% seasonal discount applies to this amount.
40,000 × .
= ₹38,000 — option (d).
67
If cosA = x, then what is the value of cotA in terms of x?
A.
B.
C.
D.2x²
Solution
cotA from cosA
sin²A + cos²A = 1, so sinA = .
cotA = .
Substituting both values.
= — option (a).
68
If cosA = and A lies in the 1st quadrant, what is the value of cotA?
A.−
B.
C.
D.
Solution
cotA from cosA
(3, 4, 5) is a Pythagorean triplet, so sinA = .
In the first quadrant all ratios are positive.
cotA = = .
= — option (b).
69
The angle of a sector is radians and the radius of the circle is 16 cm. What is the area of the sector?
A.6π cm²
B.22π cm²
C.32π cm²
D.28π cm²
Solution
Sector area
Area of a sector = r²θ, with θ in radians.
= × 16² × .
= × 256 × = 32π.
= 32π cm² — option (c).
70
If tan x = cot(2x − 10°), find the value of x.
A.50°
B.25.52°
C.40°
D.33.33°
Solution
tan = cot
tanA = cot(90° − A), so the two angles must be complementary.
x + (2x − 10°) = 90°.
3x = 100°.
x = 33.33° — option (d).
71
Two circles have radii of 16 cm and 6 cm. The distance between their centres is 30 cm. What is the length of a direct common tangent?
A.20 cm
B.22 cm
C.20 cm
D.12 cm
Solution
Direct common tangent
Direct common tangent = .
= = .
= .
= 20 cm — option (c).
72
A line drawn from the centre of a circle is perpendicular to a chord. What does it do to the chord?
A.Divides the chord into two equal parts
B.Doubles the length of the chord
C.Makes the chord tangent to the circle
D.Reduces the chord length to half
Solution
Perpendicular from centre
The perpendicular drawn from the centre of a circle to a chord always BISECTS it.
This follows from the two right triangles formed being congruent (RHS).
It divides the chord into two equal parts; it does not change its length.
Hence option (a).
73
A chord divides a circle into two arcs. The angle subtended by the major arc at the circumference is 90°. What is the central angle subtended by the major arc?
A.180°
B.110°
C.270°
D.150°
Solution
Central angle
The angle at the centre is twice the angle at the circumference for the same arc.
Angle at the circumference = 90°.
Central angle = 2 × 90°.
= 180° — option (a).
74
In a circle, chord AB subtends an angle of 85° at a point C on the circle. What is the angle subtended by the same chord AB at another point D on the same arc as C?
A.55°
B.50°
C.85°
D.40°
Solution
Angles in the same segment
Angles in the same segment of a circle are equal.
C and D lie on the same arc, so both look at chord AB from the same segment.
Hence ∠ADB = ∠ACB.
= 85° — option (c).
75
From an outside point P, two tangents PA and PB are drawn to a circle. A third tangent to the circle at a point Q intersects PA and PB at R and S respectively. If PA = 20 cm, what is the perimeter of △PRS?
A.40 cm
B.25 cm
C.20 cm
D.10 cm
Solution
Tangents from a point
Tangents drawn from an external point are equal, so RA = RQ and SB = SQ.
Perimeter = PR + RS + SP = PR + (RQ + QS) + SP.
Replacing RQ by RA and QS by SB: = (PR + RA) + (SB + SP) = PA + PB.
= 20 + 20 = 40 cm — option (a).
Finished reading? Now test yourself.
The same 25 questions, but with a timer and the answers hidden — 15 minutes.
Under compound interest the multiple grows in powers.
9 = 3², so the tripling must happen twice over.
Time = 2 × 10.
= 20 years — option (a).
From 6:00 a.m. to 2:30 p.m. is 8 hours 30 minutes = 8.5 hours.
Distance = speed × time.
= 120 × 8.5.
= 1020 km — option (c).
Sphere volume = πr³ = π(1728) = 2304π m³.
Cylinder volume = πr²h = π(4)(24) = 96π m³.
Number = .
= 24 — option (a).
Cube volume = 4³ = 64 cm³.
The largest sphere has diameter equal to the edge, so r = 2 cm.
Sphere volume = π(8) = cm³.
Empty space = 64 − cm³ — option (a).
A’s effective contribution = 3,00,000 + 1,00,000 = ₹4,00,000.
B’s contribution = ₹4,00,000.
So the ratio is 4,00,000 : 4,00,000 = 1 : 1.
A’s share = = ₹1,20,000 — option (b).
Area of a trapezium = × (sum of parallel sides) × height.
= × (10 + 6) × 8 = 64 m².
Cost = 64 × 500.
= ₹32,000 — option (a).
T₁ = × (35 + 25) × 10 = 300 m².
T₂ = × (25 + 15) × 10 = 200 m².
Total area = 300 + 200 = 500 m².
Cost = 500 × 30 = ₹15,000 — option (a).
For similar figures, area varies as the SQUARE of the corresponding length.
So the side ratio = .
= : .
= 4 : 5 — option (b).
Cost of 4 shirts = 4 × 1,500 = ₹6,000.
He receives the price of only 3 = 3 × 1,500 = ₹4,500.
Loss = 6,000 − 4,500 = ₹1,500.
Loss% = × 100 = 25% — option (b).
A fall of y% followed by a rise of y% leaves (1 − )(1 + ) = 1 − .
This must equal 0.9, so = 0.1.
y² = 1000.
y = ≈ 31.6 — option (a).
Total of the 6 numbers = 6 × 120 = 720.
Removing 18 leaves 720 − 18 = 702.
Average of the remaining five = .
= 140.4 — option (b).
For any prism, lateral surface area = perimeter of base × height.
768 = perimeter × 24.
Perimeter = .
= 32 cm — option (d).
Total of all 50 = 50 × 136 = 6800.
Top 5 total = 5 × 160 = 800; lowest 5 total = 5 × 100 = 500.
Remaining 40 total = 6800 − 800 − 500 = 5500.
Average = = 137.5 — option (b).
After the 20% discount: 50,000 × = ₹40,000.
The 5% seasonal discount applies to this amount.
40,000 × .
= ₹38,000 — option (d).
(3, 4, 5) is a Pythagorean triplet, so sinA = .
In the first quadrant all ratios are positive.
cotA = = .
= — option (b).
Area of a sector = r²θ, with θ in radians.
= × 16² × .
= × 256 × = 32π.
= 32π cm² — option (c).
Direct common tangent = .
= = .
= .
= 20 cm — option (c).
The perpendicular drawn from the centre of a circle to a chord always BISECTS it.
This follows from the two right triangles formed being congruent (RHS).
It divides the chord into two equal parts; it does not change its length.
Hence option (a).
The angle at the centre is twice the angle at the circumference for the same arc.
Angle at the circumference = 90°.
Central angle = 2 × 90°.
= 180° — option (a).
Angles in the same segment of a circle are equal.
C and D lie on the same arc, so both look at chord AB from the same segment.
Hence ∠ADB = ∠ACB.
= 85° — option (c).
Tangents drawn from an external point are equal, so RA = RQ and SB = SQ.
Perimeter = PR + RS + SP = PR + (RQ + QS) + SP.
Replacing RQ by RA and QS by SB: = (PR + RA) + (SB + SP) = PA + PB.
= 20 + 20 = 40 cm — option (a).